Components of the space of immersions 2-manifold into $\mathbb R^3$ Let $M$ be a $2$-sphere with $g$ handles. Consider the space of maps $M\to \mathbb R^3$, which are immersions [i.e. smooth maps with nondegenerate differential in each point $x\in M$], with compact-open topology. It is well-known that for $g=0$ this space is path-connected; and how about the same question when $g>0$? Is it an open question, or it's possible to find the article with the proof?
 A: The space of immersions has $2^{2g}$ components.
Let's go back to the proof of the fact for $g=0$: Smale-Hirsch immersion theory. The form of the result I want is from here.
Theorem: The space of immersions $\Sigma_g \to \Bbb R^3$ is homotopy equivalent to the space of bundle injections $T\Sigma_g \to \bf{3}$, the trivial bundle of rank 3 over $\Sigma_g$.
Instead of re-writing the wheel, I'll take the answer there starting at the cited theorem and say what must be modified.
What the author there obtains is that that space of immersions is homotopy equivalent to $\text{Maps}_*(\Sigma_g,O(3)) \times SO(3)$. Nothing about that was special to $g=0$. (Note that because we're basepointed we may as well call the mapping space $\text{Maps}_*(\Sigma_g,SO(3))$.) The issue is in his identification of that mapping space, he realizes that for $g=0$ it's $\Omega^2 SO(3)$, which is easily identified. We can't quite do this.
1) There are many homotopy classes, depending on the choice of homomorphism $\pi_1(\Sigma_g) \to \pi_1(SO(3))$ we pick. There are $2^{2g}$ many of them. If we fix the induced map to be zero, passing to the double cover of $SO(3)$ shows that this map is null, so there are precisely that many homotopy classes.
Phrasing this in terms of geometry, by picking a single immersion into $\Bbb R^3$, we've chosen a framing of the stable tangent bundle $T\Sigma_g \oplus \bf 1$. Any other immersion's framing differs by a map $\Sigma_g \to SO(3)$, and this space has $2^{2g}$ components.
Note that Smale-Hirsch immersion theory is for $C^\infty$ things, and I bet you can get it down to $C^2$. I don't know about $C^0$, $C^1$. I bet it's disconnected then too.
